National 5 Course Notes
Scientific Notation (or Standard Form)
This is a different way of writing very large and very small numbers in the form:a x 10n
where a is between 1 and 10 and n is an integer (positive or negative whole number).
Example:- Write these numbers in scientific notation.
(a) 2340000
= 2.34 x 10
(b) 0.0004485
6
= 4.485 x 10-4
Rule: Put a decimal point after the first non-zero to get a, then count the number of
places the decimal point has moved to get n.
Example:- Write these numbers in full.
(a) 6.3 x 104
= 63000
(b) 8.87 x 10-2
= 0.087
Volume of Solids
Remember to use cubic units for volume, i.e. mm3, cm3, m3, etc.
Special Prisms
V = (½bh) x h
= (½ x 6 x 11) x 14
= 33 x 14 = 462 cm3
You need to know all of the above formulae but the next ones are given in the formula
sheet.
The Sphere
Example:-
The Cone
1
V = 3 𝜋𝑟 2 ℎ
Example:- Find the volume of this cone.
V = 3.14 x 152 x 40 ÷ 3
V = 9420 cm3
Remember to use the radius and not
the diameter! 30 ÷ 2 = 15
Composite Volumes
Split the shape up into separate simple shapes then add/subtract to find volume of
composite shape.
Example
Vcone = 3.14 x 62 x 18 ÷3
Vcylinder = 3.14 x 62 x 30
Vhemisphere = (4 x 3.14 x 63 ÷3) ÷ 2
Vcone = 678.24cm3
Vcylinder = 3391.2cm3
Vhemisphere = 452.16cm3
Total Volume = 678.24 + 3391.2 + 452.16 = 4521.6cm3
Liquid Volumes
Remember:-
1 cm3 = 1 millilitre (ml)
1000 ml = 1 litre
Changing the Subject of a Formula
Simultaneous Equations
The Straight Line
The general form of the straight line is y = mx + c, where m is the gradient and c is the
y-intercept (where the line cuts the y-axis).
The formula for the gradient is 𝑚 =
𝑦2 − 𝑦1
.
𝑥 2 − 𝑥1
Pick 2 points on the line (x1, y1) and (x2, y2)
and substitute into the equation, then solve.
Example – Find the gradient of a straight line passing through the points (-3, 1) and
(2,11).
(x2, y2)
(x1, y1)
Solution:- Use 𝑚 =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
and substitute values in.
11−1
So, m = 2−(−3) =
10
5
= 2.
Example – A straight line has equation y = 5x – 3. Write down the gradient and yintercept of the line.
Solution:- Gradient – m = 5 and y-intercept – c = -3.
If you know the gradient and one point on a straight line, you can work out the equation
of the line using the formula:- y – b = m(x – a).
Example:- Find the equation of the straight line with gradient of 3, where the point
(2, 4) lies on the line.
(a, b)
Solution:- y – b = m(x – a)
=> y – 4 = 3(x – 2)
m
=> y – 4 = 3x – 6
=> y = 3x – 6 + 4
=> y = 3x – 2
Example:- Find the equation of this line.
Solution -
Pick 2 points on the line
(1,3) and (3,7)
Calculate the gradient
m = 3−1 = 2 = 2.
Pick a 3rd point
(2,5)
Sub into y – b = m(x – a)
y – 5 = 2(x – 2)
7−3
4
y – 5 = 2x – 4
y = 2x – 4 + 5
y = 2x + 1
Example:- Write down the gradient and y-intercept of 2y + 6x – 2 = 0.
Solution – Re-arrange equation into the form y =
2y + 6x – 2 = 0
2y + 6x = 2
2y = -6x + 2
y = -3x + 1
mx + c, so
take the -2 to the other side
take the 6x to the other side
divide everything by 2
So m = -3 and c = 1.
N.B.
Lines with a positive gradient slope upwards from left to right.
Lines with a negative gradient slope downwards from left to right.
Parallel lines have equal gradients.
Horizontal lines have a gradient of 0 and equation y = the y-intercept.
Vertical lines have an undefined gradient and equation x = the x-intercept.
Circles
You need to know the following formulae: d = 2r
 C = 𝜋d
 A = 𝜋r2
𝐶
 d=𝜋

𝐴
r = √𝜋
where d = diameter, r = radius, 𝜋 = 3.14, C = circumference and A = area.
Length of an Arc
arc length =
𝑎𝑛𝑔𝑙𝑒
360
x 𝜋d
Example – Calculate the length of the arc.
45
Solution:- arc length = 360 x 3.14 x 10 = 3.925cm
Remember to use the diameter
d = 2r = 2 x 5 = 10cm
Example – Find the angle at the centre of the circle.
Solution:-
Find the circumference of the circle:
C = 𝜋𝑑 = 3.14 x 14 = 43.96cm
Re-arrange the arc length formula to give:
𝑎𝑟𝑐 𝑙𝑒𝑛𝑔𝑡ℎ
angle = 𝑐𝑖𝑟𝑐𝑢𝑚𝑓𝑒𝑟𝑒𝑛𝑐𝑒 x 360o.
18
= 43.96 x 360 = 147.406733… = 147.4o.
Area of a Sector
sector area = =
𝑎𝑛𝑔𝑙𝑒
360
x 𝜋r2
Example:- Find the area of the sector.
sector area
135
= 360 x 3.14 x 5.52
= 35.619375
= 35.6cm2
Example:- Find the size of the angle at the centre of the circle.
Solution -
Find the area of the circle:
A = 3.14 x 402 = 5024m2
Re-arrange the sector formula:
angle =
𝑠𝑒𝑐𝑡𝑜𝑟 𝑎𝑟𝑒𝑎
𝑐𝑖𝑟𝑐𝑙𝑒 𝑎𝑟𝑒𝑎
x 360o
4200
Exam Type Questions
Area =
110
360
= 5024 x 360o = 300.955… = 301o
x 402 = 488.8888… = 488.9cm2
You are being asked to calculate the length of the major arc.
250
Arc length = 360 x 80 = 55.5555… = 55.6cm
Solution
QC = radius = 15cm
AC = AB – BC = 27 – 15 = 12cm
Use Pythagoras to find AQ
AQ2
= QC2 – AC2
= 152 – 122 = 225 – 144 = 81
AQ
Algebra
= √81 = 9cm so PQ = 2 x 9 = 18cm
Factorising
You must factorise in the following order:1. Common Factor – e.g. 10ab – 25b2 = 5b(2a – 5b)
N.B. always remove the Highest Common Factor
2. Difference of Two Squares – e.g. 9x2 – 25 = (3x – 5)(3x + 5)
Look for 2 terms that you can take the square root of and a subtraction sign.
3. Trinomial/Quadratic – e.g. x2 – 10x + 24 = (x – 4)(x – 6)
Look for 2 numbers which add to the middle number (-10) and multiply to the
end number (24).
Examples:1. ab2 – ac2 = a(b2 – c2) = a(b – c)(b + c)
2
common factor then diff of 2 squares
2
2. 7x + 28x + 28 = 7(x + 4x + 4) = 7(x + 2)(x + 2) = 7(x + 2)2 common factor then
trinomial
3. 3x2 – 5x – 2 = (3x + 1)(x – 2).
Method – 3 x -2 = -6 -> factors which add to -5 are -6 and 1
Re-write 3x2 – 5x – 2 as 3x2 – 6x - x – 2
Split it into 3x2 – 6x and –x – 2
Factorise each part by removing a common factor 3x2 – 6x = 3x(x-2)
and –x - 2 = 1(x - 2).
So brackets become (3x + 1)(x – 2).
Put 2 common factors of 3x
Both brackets should match after you
and 1 together.
remove common factor!
Equations and Inequalities
Remember the rule:- CHANGE SIDE, CHANGE SIGN!
All the letter terms go on the left, numbers on the right.
Examples
1. 3x + 5 = 2x + 3
=>3x – 2x = 3 – 5
=> x = -2
2. 5x + 4 = 2x + 10
=> 5x – 2x = 10 – 4
=> 3x = 6
=> x = 2
3. 11x - 20 = 10x – 15
=> 11x – 10x = -15 + 20
=> x = 5
4. 6 + x = 9 - 5x
=> x + 5x = 9 – 6
=> 6x = 3
=> x = ½
Solve inequations in the same way, but remember if you divide by a negative, you need
to change the direction of the inequality sign.
Examples
1. 7x + 3 > 3x + 29
=> 7x – 3x > 29 – 3
=> 4x > 26
=> x > 6.5
2. x – 1 < 9x – 57
=> x – 9x < -57 + 1
=> -8x < -56
=> x > 7
Pythagoras
Example
-56 ÷ -8 = 7. Change
direction of inequality sign
as dividing by a negative
Similarity
Method:-
𝑛𝑒𝑤
1. Find the scale factor ( 𝑜𝑙𝑑 )
2. Find the area factor by squaring the scale factor or volume factor by
cubing the scale factor.
3. Find the missing value.
Examples
1. Find the area of the larger rectangle.
Scale factor =
20
4
=5
Area factor = 52 = 25
Area = 25 x 12 = 300cm2
2. Find the volume of the bigger cylinder.
Scale Factor =
16
8
=2
Volume Factor = 23 = 8
Volume = 8 x 250 = 2000ml
Percentages - Compound Interest
To get the multiplier
Increase - 1 + interest rate, e.g. 8% increase = 1 + 0.08 = 1.08
Decrease - 1 - interest rate, e.g. 8% decrease = 1 - 0.08 = 0.92
Example – Sarah puts £300 in the bank and leaves it in for 5 years. The interest rate
she receives is 2.3%. How much does she have in the bank after 5 years?
Method -
amount x (1 + %)n, where n is the number of years.
Solution -
300 x (1 + 0.023)5 = 300 x 1.0235 = 336.1239227 = £336.12
Appreciation/Depreciation
This is where the value of something increases (appreciation) or decreases
(depreciation).
Example – Depreciation - Tim bought a computer three years ago for £ 1400.
Each year since then it has depreciated by 15 % of its value at the start of the year.
How much is the computer worth now?
Solution – 1400 x (1 – 0.15)3 = 1400 x 0.853 = 859.775 = £859.78
Example – Appreciation – Zoe bought a flat in 2000 for £50000. It increased in value
by 7% each year until she sold it in 2010. How much did she sell it for (to 3 significant
figures)?
Solution – 50000 x 1.0710 = 98357.56786 = £98400
Reverse Percentages
Method
1. What % are we dealing with now?
2. Find what 1% is worth
3. Now multiply by 100 to find out the original 100%
Example
Since Jim bought his house, it has appreciated in value by 12% and is now
worth £56 000.
How much did Jim pay for his house originally?
Solution
1. 112% -> 56000
2. 1% -> 56000 ÷ 112 = 500
3. 100% -> 500 x 100 = £50000
Fractions
Changing Top Heavy Fractions to Mixed Numbers – divide numerator by denominator.
Write the remainder as the new numerator.
25
4
Example –
Change
to a mixed number.
Solution -
25 ÷ 4 = 6 r 1. So
25
4
1
= 64
Changing Mixed Numbers to Top Heavy Fractions – multiply whole number by
denominator then add answer to numerator.
4
9
Example -
Change 7
to a top heavy fraction.
Solution -
7 x 9 + 4 = 63 + 4 = 67. So 7
4
9
=
67
9
Adding/Subtracting Fractions
The denominators have to be the same before you can add/subtract fractions. You will
need to change one or both fractions to have a common denominator.
Examples
3
5
1. 4 + 9
=
27
36
+
47
=36
11
=1 36
20
36
Multiply top and
bottom by same
number (usually
denominator of
other fraction).
1
1
2. 6 3 − 2 8
=4
8
24
5
= 4 24
−
3
24
1
Borrow from
whole number
so subtraction
is possible!
7
3. 5 4 − 10
=5
10
40
50
−
28
40
28
= 4 40 − 40
22
11
= 4 40 = 4 20
Remember to simplify answers as far as possible, i.e. simplify fractions and change top
heavy fractions to mixed numbers.
Multiplying Fractions
Rule – Top x Top, Bottom x Bottom. Change mixed numbers to top heavy fractions being
doing multiplication then change back to mixed number at the end.
Examples
2
3
1
1. 5 x 8
2×3
= 5×8
6
3
2. 3 4 x 5 7
=
3
= 40 = 20
=
13
38
× 7
4
13×38
4×7
=
494
28
18
9
= 17 28 = 17 14
Dividing Fractions
Rule – Similar to Multiplying Fractions but flip the second fraction, then multiply as
above.
Examples
1. ÷
5
6
2
3
2. ÷ 1
=
5
6
3
2
=
2
5
÷
=
5×3
6×2
=
2
5
× 11
=
15
12
=
20
55
×
2
5
3
1
= 1 12 = 1 4
1
10
11
10
10
=
4
11
Statistics
Mean =
𝑎𝑑𝑑 𝑢𝑝 𝑎𝑙𝑙 𝑡ℎ𝑒 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
𝑛𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑛𝑢𝑚𝑏𝑒𝑟𝑠
Median = the middle number in an ordered dataset
Mode = the most common number
Range = highest – lowest value
Example – Find the mean, median, mode and range of the following dataset.
23, 56, 24, 23, 12, 28.
Solution –
Mean =
23+56+24+23+12+28
6
=
166
6
= 27.7 (to 1 dp)
Median = 23.5 (put data in order - 12, 23, 23, 24, 28, 56)
Mode = 23
Range = 56 – 12 = 44.
Five-Figure Summary
The five figure summary consists of the following
 L = the lowest figure in the data set
 Q1 = quartile 1 (median of lower half of dataset)
 Q2 = quartile 2 which is the same as the median
 Q3 = quartile 3 (median of upper half of dataset)
 H = the highest figure
Example – What is the 5-figure summary for this dataset?
2 4 5 5 6 7 7 8 9 10
L Q 1 Q2
Q3 H
L = 2, Q1 = 5, Q2 = 6.5, Q3 = 8, H = 10
Semi-Interquartile Range (SIQR)
SIQR =
𝑄3 − 𝑄1
2
=
8−5
2
= 1.5
Standard Deviation
This is a measure of spread. The lower the standard deviation, the less spread out the
scores/data points are.
Scattergraphs
Example – Draw a line of best fit onto the scattergraph and determine its equation.
Solution -
Try to get your line to go through as
many points as possible, with as
many points above as below the line.
Now find the equation of the line:



c = 10 (read off the graph)
Pick 2 points on the line (must
be on the graph too), e.g. (0,10)
and (60, 60).
Calculate gradient
60−10
5
m = 60−0 = 6.
5
6
So y = x + 10.
Estimate the Maths mark of a student who got 70% in Physics.
Solution – Put x = 70 into equation => y = 5 x 70 ÷ 6 + 10 = 68.333… = 68%.